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The graph of $y=x^x$ looks like this:

Graph of y=x^x.

As we can see, the graph has a minimum value at a turning point. According to WolframAlpha, this point is at $x=1/e$.

I know that $e$ is the number for exponential growth and $\frac{d}{dx}e^x=e^x$, but these ideas seem unrelated to the fact that the mimum value of $x^x$ is $1/e$. Is this just pure coincidence, or could someone provide an intuitive explanation (i.e. more than just a proof) of why this is?

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    $\begingroup$ I would appeal to the logarithmic derivative of the function to show that. $\endgroup$ – Triatticus Jan 4 '16 at 20:25
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Hint: Note that $x^x = e^{x\log x}$.

So minimizing $x^x$ is the same as minimizing $x\log x$

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Let $f(x)=x^x$. Note that $f(x)$ is only defined for $x>0$.

Then $$ \ln f(x)=x\cdot\ln x\tag{1} $$ Differentiating (1) gives $$ \frac{1}{f(x)}f^\prime(x)=x\frac{1}{x}+\ln x=1+\ln x $$ Note that we have used the chain rule and the product rule.

Solving for $f^\prime(x)$ gives $$ f^\prime(x)=f(x)(1+\ln x)=x^x(1+\ln x) $$ Can you use this to locate the critical points of $f(x)$?

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    $\begingroup$ I think you can leave the absolute value off since $x>0$ on its domain, right? $\endgroup$ – Alejandro Jan 4 '16 at 20:31
  • $\begingroup$ @Subjunctive sure $\endgroup$ – Brian Fitzpatrick Jan 4 '16 at 20:31
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Hint: actually you are looking for a local/global minimum .. so look at the derivative of the function $f(x) = x^x$ $$f'(x) = x^x (\log (x)+1)$$ which equals $0 \iff x = \frac{1}{e}$

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Try to minimize logarithm of $ y= x^x,$ i.e., $y=x \, \log x$

Its derivative is

$$ 1 + \log(x) $$

When equated to zero, it solves to that the minimum of $y(x)$ ocuurs at

$$ x= \dfrac{1}{e}= 0.36788$$

and the minimum value is

$$ y_{min}= \dfrac{1}{{e}^{\frac{1}{e}}} \approx 0.6922$$

The tiny red dot shown in your graph for minimum is:

$$ (x,y) = (0.36788, 0.6922) $$

Note that the minimum value is not $\dfrac{1}{e}\, ! \, $ ... but is the reciprocal of $e^{th} $root of $e.$

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